hdu 2612 Find a way

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=2612
两次bfs，注意若有一方无法到达某个kfc，这个点就不能取。。

#include<algorithm>#include<iostream>#include<cstdlib>#include<cstring>#include<cstdio>#include<vector>#include<queue>#include<map>using std::min;using std::cin;using std::cout;using std::endl;using std::find;using std::sort;using std::map;using std::pair;using std::queue;using std::vector;using std::multimap;#define pb(e) push_back(e)#define sz(c) (int)(c).size()#define mp(a, b) make_pair(a, b)#define all(c) (c).begin(), (c).end()#define iter(c) decltype((c).begin())#define cls(arr,val) memset(arr,val,sizeof(arr))#define cpresent(c, e) (find(all(c), (e)) != (c).end())#define rep(i, n) for (int i = 0; i < (int)(n); i++)#define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i)const int N = 210;const int INF = ~0u >> 1;typedef unsigned long long ull;char G[N][N];int H, W, Sx, Sy, Dx, Dy, dist[2][N][N];const int dx[] = { 0, 0, -1, 1 }, dy[] = { -1, 1, 0, 0 };struct Node {    int x, y;    Node(int i = 0, int j = 0) :x(i), y(j) {}}Y, M;void bfs(int x, int y, bool d) {    queue<Node> q;    q.push(Node(x, y));    dist[d][x][y] = 0;    while (!q.empty()) {        Node t = q.front(); q.pop();        rep(i, 4) {            int nx = t.x + dx[i], ny = t.y + dy[i];            if (nx < 0 || nx >= H || ny < 0 || ny >= W) continue;            if (G[nx][ny] == '#' || dist[d][nx][ny]) continue;            dist[d][nx][ny] = dist[d][t.x][t.y] + 1;            q.push(Node(nx, ny));        }    }}int main() {#ifdef LOCAL    freopen("in.txt", "r", stdin);    freopen("out.txt", "w+", stdout);#endif    while (~scanf("%d %d", &H, &W)) {        cls(dist, 0);        vector<Node> kfc;        rep(i, H) {            scanf("%s", G[i]);            rep(j, W) {                if (G[i][j] == 'Y') Sx = i, Sy = j;                if (G[i][j] == 'M') Dx = i, Dy = j;                if (G[i][j] == '@') kfc.pb(Node(i, j));            }        }        bfs(Sx, Sy, 0), bfs(Dx, Dy, 1);        int res = INF;        rep(i, sz(kfc)) {            Node &k = kfc[i];            int A = dist[0][k.x][k.y], B = dist[1][k.x][k.y];            if (!A || !B) continue;            res = min(res, A + B);        }        printf("%d\n", res * 11);    }    return 0;}